Question

The weekly incomes of a large group of middle managers are normally distributed with a mean of $1,800 and a standard deviation of $200. Use the normal distribution to calculate the following:

A. What percent of income lies within $400 of the population mean?

B. 68% of observations lie within which two values?

Answer 1

A) P(1400 < X < 2200)

= P((1400 - )/ < (X - )/ < (2200 - )/)

= P((1400 - 1800)/200 < Z < (2200 - 1800)/200)

= P(-2 < Z < 2)

= P(Z < 2) - P(Z < -2)

= 0.9772 - 0.0228

= 0.9544

b) P(-x < X < x) = 0.68

or, P(-z < Z < z) = 0.68

or, P(Z < z) - P(Z < -z) = 0.68

or, P(Z < z) - (1 - P(Z < z)) = 0.68

or, P(Z < z) - 1 + P(Z < z) = 0.68

or, 2P(Z < z) = 1.68

or, P(Z < z) = 0.84

or, z = 0.99

or, (X - 1800)/200 = 0.99

or, x = 0.99 * 200 + 1800

or, x = 1998

The other value is = 1800 - (1998 - 1800) = 1602